Chapter 4, Problem 142P

(a)

To determine

Check whether the puck move or not due to $5.0\text{N}$ force in forward direction.

Answer to Problem 142P

Puck will not move.

Explanation of Solution

The mass of puck is $2.0\text{kg}$, coefficient of static friction between the board and puck is $0.35$, coefficient of kinetic friction between the board and puck is $0.25$, and the applied force is $5.0\text{N}$.

The puck will move or not depends on whether the applied force is greater than or lesser than that of the frictional force between the puck and the board. The puck will move if the applied force is greater than that of the frictional force.

Write the expression to calculate the frictional force between the puck and the board.

$f_{s,\max }={\mu }_{s}N$

Here, the maximum value of static frictional force between the puck and the board is $f_{s,\max }$, coefficient of static friction between the board and puck is ${\mu }_s$, and the normal reaction force on puck is $N$.

Write the equation for $N$.

$N=mg$

Here, the mass of puck is $m$ and the gravitational acceleration of earth is $g$.

Rewrite the equation for $f_{s,\max }$ by substituting the above relation for $N$.

$f_{s,\max }={\mu }_{s}m{g}_{earth}$

Conclusion:

Substitute $0.35$ for ${\mu }_s$, $2.0\text{kg}$ for $m$, and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to find $f_{s,\max }$.

$\begin{array}{c}{f}_{s,\max }=\left(0.35\right)\left(2.0\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =6.86\text{N}\end{array}$

It is found that $f_{s,\max }$ is lesser than that of the applied force $5.0\text{N}$.

Therefore, the puck cannot move with $5.0\text{N}$ force.

(b)

To determine

Check whether the puck move or not due to $7.5\text{N}$ force in forward direction.

Answer to Problem 142P

Puck will move.

Explanation of Solution

The mass of puck is $2.0\text{kg}$, coefficient of static friction between the board and puck is $0.35$, coefficient of kinetic friction between the board and puck is $0.25$, and the applied force is $5.0\text{N}$.

The puck will move or not depends on whether the applied force is greater than or lesser than that of the frictional force between the puck and the board. The puck will move if the applied force is greater than that of the frictional force.

Write the condition for the puck to move

$F_{applied}>f_{s,\max }$

Here, the applied force is $F_{applied}$.

Conclusion:

Substitute $7.5\text{N}$ for $F_{applied}$ and $6.86\text{N}$ for $f_{s,\max }$ in the above condition to check whether it is valid or not.

$7.5\text{N}>6.86\text{N}$

It is found that $F_{applied}>f_{s,\max }$.

Therefore, the puck cannot move with $5.0\text{N}$ force.

(c)

To determine

Check whether the puck move or not due to $7.5\text{N}$ force in forward direction.

Answer to Problem 142P

Puck will move.

Explanation of Solution

The mass of puck is $2.0\text{kg}$, coefficient of static friction between the board and puck is $0.35$, coefficient of kinetic friction between the board and puck is $0.25$, and the push force is $6.0\text{N}$.

Write the equation for net force on the puck.

$ma=F_{push}-{\mu }_{k}g$

Here, the acceleration of puck is $a$, coefficient of kinetic friction between the board and puck is ${\mu }_k$, and the push force is $F_{push}$.

Conclusion:

Substitute $2.0\text{kg}$ for $m$, $6.0\text{N}$ for $F_{push}$, $0.25$ for ${\mu }_k$, and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation to find $a$.

$\begin{array}{c}\left(2.0\text{kg}\right)a=6.0\text{N}-\left(\left(6.0\text{N}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\right)\\ =\frac{6.0\text{N}-58.8\text{N}}{2.0\text{kg}}\\ =0.6\text{N}/{\text{m}}^{\text{2}}\end{array}$

Therefore, the puck can move.

(d)

To determine

Whether the acceleration of puck will be same, lesser than, or greater than that of in earth for the same applied force.

Answer to Problem 142P

Acceleration of puck will be greater in moon that in earth.

Explanation of Solution

The mass of puck is $2.0\text{kg}$, coefficient of static friction between the board and puck is $0.35$, coefficient of kinetic friction between the board and puck is $0.25$, and the applied force is $6.0\text{N}$.

Write the equation for net force on the puck.

$ma=F_{push}-{\mu }_{k}g$

Rewrite the above relation in terms of $a$.

$a=\frac{F_{push}}{m}-{\mu }_{k}g$

From the above equation it can be seen that the acceleration of puck depends on the gravitational acceleration exerted on puck.

Rewrite the above equation for puck in moon.

$a_{moon}=\frac{F_{push}}{m}-{\mu }_k{g}_{moon}$

Here, the acceleration of puck in moon is $a_{moon}$ and the gravitational acceleration of moon is $g_{moon}$.

The value of $g_{moon}$ is lesser than that of $g$.This means that $\frac{F_{push}}{m}-{\mu }_k{g}_{moon}$ is greater than that of $\frac{F_{push}}{m}-{\mu }_{k}g$.It means that the fractional force on puck in moon is less than that in earth. Thus, $a_{moon}$ will be greater than that of $a$.

Conclusion:

Therefore, the acceleration of puck will be greater in moon that in earth.